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e x 1 + x 2 - Mathematics

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\[\frac{e^x}{1 + x^2}\] 

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Solution

\[\text{ Let } u = e^x ; v = 1 + x^2 \]
\[\text{ Then }, u' = e^x ; v' = 2x\]
\[\text{ Using the chain rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{e^x}{1 + x^2} \right) = \frac{\left( 1 + x^2 \right) e^x - e^x \left( 2x \right)}{\left( 1 + x^2 \right)^2}\]
\[ = \frac{e^x + x^2 e^x - 2x e^x}{\left( 1 + x^2 \right)^2}\]
\[ = \frac{e^x \left( 1 + x^2 - 2x \right)}{\left( 1 + x^2 \right)^2}\]
\[ = \frac{e^x \left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2}\]
\[\]

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.5 | Q 8 | Page 44

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